What is the inverse of #h(x)=-4/x^2#?

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The furthest I got on this question was #y=sqrt(-1/4x#

But I don't think that's right

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Answer 1 · 2016-10-23T14:43:29

Do you want + or -?

Exchange x and y.

#x = -4/y^2#

Multiply by #y^2#

#xy^2 = -4#

Divide by x, then take the root.

#y^2 = -4/x#

#y = ± sqrt{-4/x} = h^-1(x)#